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I haven't done too many proofs, but I'd like to attempt more and thought I'd take a shot at this one, which is from Penrose's book "The Road To Reality."

The problem is to prove that the function given by $y = 0, x \leq 0$ and $y = e^{-1/x}, x > 0$ is $C^\infty$ smooth.

When x is less than 0, there's no problem as obviously the function is infinitely differentiable. Similarly, when x is greater than zero the function is infinitely differentiable, by the properties of the exponential function. As I see it, the difficulty arises in what happens right at 0. For the function to be differentiable there, the function needs to be continuous at 0, and for that to happen both the right hand and left hand limits need to equal 0. So for the function to be infinitely differentiable, one would need to show that in the limit the function $e^{-1/x}$, and all its derivatives, go to zero as x goes to 0.

Am I on the right track? Thanks for any advice.

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Yes, you are on the right track... –  Fabian May 5 '11 at 18:52
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up vote 5 down vote accepted

Good start! Here's a little hint: you don't have to compute the $n$th derivative of $e^{-1/x}$ explicitly (that would be a mess). It's enough to show what structure it has, namely "a polynomial in $1/x$, times $e^{-1/x}$". This will allow you to deduce that it tends to zero as $x\to 0^+$.

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Thanks. I came to a similar conclusion a few minutes ago before I checked for replies, honest! Since the limit of a product is the product of a limit, all that needs to be done to complete the proof show that $1/{x^{(n+1)}}$ tends to zero for all positive integers n, where n is the number of differentiations performed on the original equation. Easy do do via induction. –  Bitrex May 5 '11 at 19:17
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Hm, I'm not sure I understand your notation. Just to avoid misunderstandings: $1/x \to +\infty$ as $x\to 0^+$, so you will not have a limit as simple as $0 \cdot 0$, but the point is that $e^{-1/x}$ tends to zero much faster than the polynomial in $1/x$ grows. –  Hans Lundmark May 5 '11 at 20:55
    
You're right, I got confused on the direction I was taking the limit there! –  Bitrex May 6 '11 at 8:36
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The derivatives are on the positive side do not seem impossible to calculate. I think they are

$$f^{(n)}(x) = e^{-1/x} \sum_{i=n+1}^{2n} {i \choose 2n-i}{i-1 \choose 2n-i}(2n-i)! (-x)^{-i}$$

as can be shown by induction. They have the limit $0$ as $x \to 0$ from above (easiest to see by first letting $y=1/x$ and then taking the limit as $y \to \infty$), so the function is infinitely differentiable everywhere.

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To show that all $f(x) \in C^\infty$, it is useful to find $f^{(n)}(x)$, because then it is easy to show the differentiability in $0$.

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